Linear Algebra (1 Viewer)

RenegadeMx · Jun 18, 2015

5a, 5bii& iii

InteGrand · Jun 18, 2015

RenegadeMx said:
5a, 5bii& iii

$5. (a) Since $A^{m}\bold{v}=\bold{0}$, all further powers are $\bold{0}$ too, i.e. $\bold{0}=A^{m+1}\bold{v}=A^{m+2}\bold{v}=\cdots$ (this is seen by pre-multiplying both sides of $A^{m}\bold{v}=\bold{0}$ by powers of $A$). Also, since $A^{m-1}\bold{v}\neq \bold{0}$, $A^{m-2}\bold{0}\neq \bold{0}$, as $\bold{0}\neq A^{m-1}\bold{v} = A\left( A^{m-2}\bold{v} \right)$ (if $A^{m-2}\bold{v}$ were $\bold{0}$, then so would $A^{m-1}\bold{v}$ be). Continuing inductively, we can see that none of the elements of $S$ can be $\bold{0}$.$

$Now, to prove the requirement, we need to show that the only solution to the equation$

$$\lambda_1 \bold{v}+\lambda_2 A\bold{v}+\lambda_3 A^2 \bold{v}+\cdots + \lambda_{m-1}A^{m-2}\bold{v} + \lambda_{m}A^{m-1}\bold{v}=\bold{0}$ (*)$

$is $\bold{\Lambda}\equiv \begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_m \end{pmatrix}=\bold{0}$.$

$To show this, pre-multiply both sides of (*) by $A^{m-1}$, then all vectors involving terms with powers of $A$ higher than $m-1$ vanish (as we showed that $\bold{0}=A^{m+1}\bold{v}=A^{m+2}\bold{v}=\cdots$), and we are left with $\lambda_1 \underbrace{A^{m-1}\bold{v}}_{\neq\bold{0}}=\bold{0}\Rightarrow \lambda_1 = 0$.$

$We basically carry on in this fashion, pre-multiplying both sides of (*) by $A^{m-j}$ in the order $j=2,3,4,...,m-1$ and we will successively see that this forces $\lambda_2 = 0,\lambda_3=0,...,\lambda_{m-1}=0$, and then finally forces $\lambda_m=0$, as required. \blacksquare$

InteGrand · Jun 18, 2015

RenegadeMx said:
5a, 5bii& iii

$For 5 (b) (ii), note that the column $j$ element of row $k$ is $(j+(k-1))^2$, where $k,j\in \{1,2,3,...,100\}$. Hence the elementary row operation $R_k \to R_k - R_l$ makes the new $R_k$ have the following column $j$ element: $(j+(k-1))^2 -(j+(l-1))^2 = (2j+(k+l-2))(k-l)$ (difference of two squares). Hence if we apply the row operations $R_{51} \to R_{51} - R_{50}$ and $R_{98}\to R_{98} - R_3$, we have the resultant matrix $A_2$ have the following column $j$ elements: $R_{98}: (2j+101-2)(98-3)$ and $R_{51}: (2j+101-2)(51-50)$. Hence $A_2$ has two rows which are scalar multiples of each other, so $\det(A_2)=0$, and $\det(A)=\det(A_2)=0$, as the determinant is unchanged by performing elementary row operations. \blacksquare$

erckle999 · Jun 21, 2015

Even though you didn't ask, my solution to (c). It seems very long. If you have a more concise answer please share it.

Couldn't get Latex working so here is a picture:

RenegadeMx · Jun 21, 2015

erckle999 said:
Even though you didn't ask, my solution to (c). It seems very long. If you have a more concise answer please share it.

Couldn't get Latex working so here is a picture:
View attachment 32131
View attachment 32132

i did, i wanted those specific questions for first pic, and all of the 2nd pic pretty much, and thanks for that

Linear Algebra (1 Viewer)

RenegadeMx

Kosovo is Serbian

InteGrand

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InteGrand

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erckle999

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RenegadeMx

Kosovo is Serbian

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